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Torque as cross product

Torque measures how a force changes angular momentum about a chosen point or axis. The same force can produce a large torque, a small torque, or zero torque depending on where and in which direction it acts.

Definition about a point

If a force \(\mathbf F\) acts at a point whose position relative to origin \(O\) is \(\mathbf r\), its torque about that origin is

\[\boldsymbol\tau_O=\mathbf r\times\mathbf F\]

Its magnitude is

\[\tau=rF\sin\phi\]

where \(\phi\) is the smaller angle between the vectors. The SI unit is newton metre. Although this unit has the same base dimensions as a joule, torque and energy are different quantities: torque is an axial vector associated with rotation, while energy is a scalar.

The right-hand rule gives torque direction. Curl the fingers of the right hand from \(\mathbf r\) toward \(\mathbf F\) through the smaller angle; the thumb points along \(\boldsymbol\tau\).

Lever arm and line of action

The perpendicular lever arm \(r_\perp\) is the shortest distance from the origin to the force's line of action. Therefore,

\[\tau=F r_\perp\]

Equivalently, the component of force perpendicular to \(\mathbf r\) produces the torque:

\[\tau=rF_\perp\]

Use one of these interpretations, not both at once. Multiplying perpendicular force by perpendicular distance again would count the sine factor twice.

A force whose line of action passes through the origin has zero lever arm and zero torque about that origin. The force can still accelerate the center of mass. Torque and net force answer different questions.

import numpy as np

origin_m = np.zeros(3)
application_m = np.array([0.4, 0.0, 0.0])
force_n = np.array([0.0, 20.0, 0.0])
torque_nm = np.cross(application_m - origin_m, force_n)
print("Torque (N m):", torque_nm)

Component calculation

For Cartesian components,

\[\mathbf r\times\mathbf F=\begin{bmatrix}r_yF_z-r_zF_y\\r_zF_x-r_xF_z\\r_xF_y-r_yF_x\end{bmatrix}\]

The order matters: \(\mathbf F\times\mathbf r=-\mathbf r\times\mathbf F\). Swapping operands produces the opposite torque.

For motion in the \(xy\), only the perpendicular component is needed:

\[\tau_z=r_xF_y-r_yF_x\]

If positive \(z\) points out of the page, positive \(\tau_z\) tends to produce counterclockwise angular acceleration under the usual convention.

Example: pushing a door

Suppose a force of \(20\,\mathrm N\) is applied perpendicularly at a point \(0.80\,\mathrm m\) from a door hinge. The torque magnitude is

\[\tau=(0.80\ \mathrm m)(20\ \mathrm N)=16\ \mathrm{N\,m}\]

Figure 1(a) shows the geometry of this calculation. The upward force's line of action misses the hinge, and its perpendicular distance from the hinge equals the door's lever length. The right-hand rule gives torque out of the page, corresponding to the curved counterclockwise arrow. Applying the same force halfway from the hinge halves this distance and therefore halves the torque.

Now follow the dashed line through the force in Figure 1(b): it passes through the hinge. The perpendicular lever arm is zero even though the handle is still far from the hinge. This is why pushing toward the hinge produces no torque about it.

Net torque from several forces

Torques about the same origin add vectorially:

\[\boldsymbol\tau_{\mathrm{net},O}=\sum_k(\mathbf r_k\times\mathbf F_k)\]

For the door in Figure 1, keep O at the hinge when adding further applied forces. Each force needs its own application point and line of action, just as the two panels use the same origin but obtain different torques from different force directions.

Every lever vector must begin at the same origin. Mixing torques calculated about different points has no clear meaning. If a force is distributed over a body, replace the sum with an integral:

\[\boldsymbol\tau_O=\int\mathbf r\times d\mathbf F\]

For uniform gravity near Earth's surface, the net gravitational torque on a rigid body can be calculated as if its weight acted at the center of mass. In a nonuniform gravitational field, this shortcut is not generally exact because tidal variation across the body matters.

forces = [
    (np.array([0.4, 0, 0]), np.array([0, 20, 0])),
    (np.array([-0.2, 0, 0]), np.array([0, -10, 0])),
]
net_torque = sum(
    (np.cross(point - origin_m, force) for point, force in forces), np.zeros(3)
)
print("Net torque (N m):", net_torque)

A force couple

Two equal and opposite forces separated by a distance have zero net force but can have nonzero net torque. This arrangement is a couple. For ideal equal-and-opposite forces, the couple torque is independent of the chosen origin because shifting the origin adds opposite terms that cancel.

A couple can change rotation without accelerating the center of mass. Turning a steering wheel with opposite tangential forces from two hands approximates this situation.

Changing the origin

Torque generally depends on origin. If origin is shifted by vector \(\mathbf a\), net torque changes as

\[\boldsymbol\tau_{O'}=\boldsymbol\tau_O-\mathbf a\times\mathbf F_{\mathrm{net}}\]

When net force is zero, net torque is independent of origin. Otherwise, always state the point about which torque is calculated. In statics, choosing a pivot strategically can make unknown reaction forces contribute zero torque, but it does not make those forces physically disappear.

Torque and angular momentum

The general rotational balance law about a fixed point in an inertial frame is

\[\boldsymbol\tau_{\mathrm{net}}=\frac{d\mathbf L}{dt}\]

This is more general than \(\tau=I\alpha\). The scalar relation applies to rotation about a fixed principal axis with constant moment of inertia, or other circumstances where the required components reduce consistently.

In unrestricted three-dimensional rigid-body motion, angular momentum and angular velocity are related through the inertia tensor:

\[\mathbf L=\mathbf I\boldsymbol\omega\]

Torque need not be parallel to angular acceleration, and angular momentum need not be parallel to angular velocity. Dividing a three-dimensional torque vector by one scalar inertia is generally incorrect.

Torque, work, and power

For fixed-axis rotation, a torque component about the axis does differential work

\[dW=\tau\,d\theta\]

and instantaneous power is

\[P=\tau\omega\]

In vector form, rotational power is \(P=\boldsymbol\tau\cdot\boldsymbol\omega\). A torque perpendicular to angular velocity changes its direction without instantaneous rotational power, a key idea in gyroscopic precession.

Verification strategy

Test geometry first. Parallel and antiparallel force and lever vectors give zero torque. Perpendicular unit vectors produce unit magnitude with the right-hand-rule direction. Reversing force reverses torque. Translating origin and application point together leaves the lever vector unchanged.

Then test dynamics separately. For a fixed positive inertia, doubling torque doubles angular acceleration. Zero net torque preserves angular momentum, not necessarily zero angular velocity. For numerical force data, compare the cross-product calculation with the perpendicular-distance calculation in nondegenerate cases.

Reliability checks

Torque is the geometry-aware form of rotational influence. The cross product preserves the origin, direction, and lever-arm information needed to connect applied forces with angular momentum and fixed-axis acceleration.

A worked Python notebook

Apply a 20 newton force at a lever arm of 0.4 metres. The question for this calculation is precise: how does torque about pivot change when we change force angle, while keeping the other choices fixed? The earlier sections explain the physical relationship. Here we turn one part of that relationship into a calculation that you can run, inspect, and change in Google Colab.

The angle is between the lever arm and force; the scalar sign refers to the chosen normal axis. Keep that sentence beside the results. A formula can return a number outside its useful range, and a plotting library will usually draw that number without asking whether the physical assumptions still hold. The selected interval is from 0 to 3.14159 rad. It gives us a specific domain to discuss instead of leaving the graph's range to a software default.

Use the notebook link above the lesson, run the setup cell, and then run the cells in order. The code below is also included there. NumPy handles the arrays of numerical values, SymPy keeps the equation symbolic, and Matplotlib draws the results. If these libraries or notebook cells are unfamiliar, start with Your first physics notebook.

State the relationship before choosing a graph

In this small example, q stands for force angle, expressed in rad. The output is torque about pivot, expressed in N m. The equation below uses the numerical values in those units. Any fixed coefficients therefore carry the units needed to make the relation consistent. A value supplied in another unit must be converted before it enters this expression.

\[f(q) = 8 \sin{\left(q \right)}\]

This is a function of one changing input. It is not a statement that the physical system has only one variable. We deliberately hold the other inputs fixed so that one cause of variation is visible. Changing several inputs at once would make it harder to tell which change produced the output difference. Write down the held-fixed choices before trying a different experiment.

At 1.5708 rad, the expression gives 8 N m. This is our reference calculation. It is a prediction of the stated example, not a new measurement. Its displayed digits help us compare implementations. They should not be read as a claim that physical inputs are known to six significant figures.

import numpy as np
import sympy as sp
import matplotlib.pyplot as plt

q = sp.Symbol("q", real=True)
expression = sp.sympify("8*sin(q)", locals={"q": q})
reference_input = 1.5707963267948966
reference_output = float(expression.subs(q, reference_input))
print("Input:", reference_input, "rad")
print("Predicted torque about pivot:", reference_output, "N m")

sp.Symbol declares a mathematical symbol. It does not assign a numerical measurement to that symbol. subs then replaces the symbol with our chosen input, and float converts the resulting symbolic number into a standard Python numerical value. Keeping these steps separate lets us inspect the equation before numerical substitution. It also prevents a later plot from becoming the only surviving record of the model.

Compare a few cases you can explain

Before drawing hundreds of points, calculate three. The lower endpoint gives 0 N m. The midpoint, at 1.5708 rad, gives 8 N m. The upper endpoint gives 9.79717e-16 N m. Read these together with the assumptions: they describe different choices of force angle within the same example, with the other inputs held fixed.

These three values are a useful first check, but they do not establish the shape between them. Two endpoints can be equal even when the curve bends or oscillates in between. Likewise, a large change between endpoints does not tell us whether the relationship is linear. We need the equation and intermediate samples to answer those questions.

evaluate = sp.lambdify(q, expression, modules="numpy")
inputs = np.array([0.0, 1.5707963267948966, 3.141592653589793])
outputs = np.asarray(evaluate(inputs), dtype=float)
outputs = np.broadcast_to(outputs, inputs.shape)
for input_value, output_value in zip(inputs, outputs):
    print(f"{input_value:.6g} rad -> {output_value:.6g} N m")
assert np.all(np.isfinite(outputs))

lambdify turns the symbolic expression into a numerical function. We explicitly select NumPy as its numerical backend so it can accept an array. The same expression now has two useful forms: one for algebra and one for calculating many cases. The assertion checks that the chosen inputs produce finite outputs. It would catch an infinity or an undefined numerical result, but a finite value can still be physically inappropriate.

Make a graph that answers the question

The horizontal axis shows force angle; the vertical axis shows torque about pivot. Every plotted point is a prediction of the same expression. We mark the reference case separately so that a reader can connect the earlier printed value to its place on the curve. The axis labels include units because a shape without a scale cannot tell us the size of a physical effect.

input_grid = np.linspace(0.0, 3.141592653589793, 301)
output_grid = np.asarray(evaluate(input_grid), dtype=float)
output_grid = np.broadcast_to(output_grid, input_grid.shape)
fig, ax = plt.subplots(figsize=(7, 4))
ax.plot(input_grid, output_grid, label="Stated model")
ax.scatter([reference_input], [reference_output], label="Reference case", zorder=3)
ax.set_xlabel("Force angle (rad)")
ax.set_ylabel("Torque about pivot (N m)")
ax.legend()
ax.grid(alpha=0.25)
fig.tight_layout()
plt.show()

Look for a sign change, a turning point, or a region where the curve becomes steep. Connect each feature to the expression rather than to the colour of the line. If the curve has a minimum or maximum, changing the input in opposite directions may have similar effects on the output near that point. If it is nearly straight over a short interval, a local linear approximation may be useful even when the complete expression is nonlinear.

Increasing the number of samples makes the line smoother and may reveal features that a coarse grid missed. It does not improve the physical inputs or enlarge the model's validity range. For quantities that can take only discrete values, use the line as a guide to the relationship and interpret only the allowed input values. A plotting function does not enforce such restrictions for you.

Measure the local sensitivity

A derivative asks a more specific question than the whole graph: how quickly does torque about pivot change near the reference input? Its units are the output unit divided by the input unit. For this calculation, the derivative at 1.5708 rad is 4.89859e-16 in those ratio units. Locally, the output increases as the input increases. This statement is local; the sign or size of the slope can be different elsewhere on the curve.

\[f'(q) = 8 \cos{\left(q \right)}\]

For a small input change, multiply the derivative by that change to estimate the output change. Then compare that estimate with a fresh evaluation of the full expression. The distinction between these two calculations matters: the derivative gives a tangent approximation, while substitution evaluates the chosen expression at the new point.

slope_expression = sp.diff(expression, q)
slope_at_reference = float(slope_expression.subs(q, reference_input))
changed_input = 1.5865042900628454
linear_prediction = reference_output + slope_at_reference * (
    changed_input - reference_input
)
full_prediction = float(evaluate(changed_input))
print("Symbolic derivative:", slope_expression)
print("Tangent prediction:", linear_prediction, "N m")
print("Full expression:", full_prediction, "N m")
print("Approximation difference:", full_prediction - linear_prediction, "N m")

Here the changed input is 1.5865 rad. The tangent estimate is 8 N m, while the full expression gives 7.99901 N m. If the two values are close, it means the chosen input change is small enough for this local approximation at this point. It does not mean the entire model is linear. If the slope is zero, a quadratic or higher-order term may supply the leading change instead.

This is also the beginning of uncertainty propagation. If the input has a small stated standard uncertainty, the magnitude of the slope times that uncertainty estimates one contribution to output uncertainty. Do not automatically use the same rule for a wide interval, a discontinuity, a turning point with zero first derivative, or several correlated uncertain inputs. Those situations need more of the model than one slope.

Check the derivative with numerical differences

We can estimate the same slope from values on either side of the reference point. Subtract the lower output from the upper output and divide by twice the input step. Parentheses are essential: the entire output difference must be divided by the entire input interval. Without them, Python can execute a different mathematical expression while still producing a plausible number.

The calculation below uses three decreasing step sizes. Each estimate is compared with the symbolic derivative, so the reference is independent of the numerical difference formula. Near an endpoint, it is acceptable here to evaluate a small extension only because the displayed mathematical expression is smooth there; such a check does not assert that extrapolated values describe additional physical cases.

input_span = 3.141592653589793
for fraction in [1e-2, 1e-3, 1e-4]:
    step = input_span * fraction
    upper = float(evaluate(reference_input + step))
    lower = float(evaluate(reference_input - step))
    numerical_slope = (upper - lower) / (2 * step)
    difference = numerical_slope - slope_at_reference
    print(f"step={step:.6g}: slope={numerical_slope:.6g}, difference={difference:.6g}")

For a smooth function, reducing the step often reduces the approximation error at first. Very small steps can eventually make subtraction of nearly equal floating-point values less reliable. Therefore “smaller” is not an unlimited guarantee of “better.” Compare the observed behaviour over a useful range of steps instead of choosing the smallest number the computer can represent.

An absolute slope difference is useful when the expected slope is near zero. A relative difference would divide by a very small reference and could look enormous even when the absolute discrepancy is tiny. Away from zero, a relative difference can help compare quantities with very different scales. The tolerance should reflect the question and the numerical method, rather than a universal decimal threshold copied into every test.

Change the example and preserve the reasoning

Make one copy of the reference calculation before editing it. Choose a second value of force angle inside the stated interval, predict whether torque about pivot will rise or fall, and then evaluate the expression. Place the prediction and the result in a text cell. If they disagree, identify whether the mistake was in the sign, the unit conversion, the algebra, or the assumption about how the curve behaves.

Next choose a shorter plotting interval around the reference point. Compare the local curve with its tangent approximation. Then widen the interval again. The difference shows why a useful local estimate need not describe the whole range. Keep the axis labels and numerical scale visible when comparing the plots; automatic axis limits can make two different effects appear equally large.

Finally, return to the physical limit stated at the start: The angle is between the lever arm and force; the scalar sign refers to the chosen normal axis. Describe one change to the setup that would violate that limit. State what new input, equation, or observation would be needed before trusting an answer in the changed situation. This last step connects the Python experiment to the physics instead of treating successful execution as the only criterion for a useful result.

Before saving, restart the runtime and run all cells in order. A result that depends on an earlier, now-deleted input cell is not reproducible from the saved notebook. Keep unrounded values for later calculations, round only the reported output, and label any new data as measured, assumed, or generated. The graph and printed values should remain traceable to the same expression and the same choices of inputs.

Recover the output change by integrating the slope

There is another way to check that the expression and its derivative describe the same relationship. Add up the small output changes predicted by the slope across the chosen input interval. In the continuous limit, this integral equals the output at the upper endpoint minus the output at the lower endpoint. The result has units of N m: the input unit in the integration cancels the input unit in the denominator of the slope.

This statement is about the mathematics of this particular relationship. It does not assign a new physical meaning to an arbitrary area under a graph. For example, an area under force versus distance represents work, while an area under a different pair of quantities may represent something else or have no standard name. Always carry the axis units through the calculation before naming an integral.

slope_function = sp.lambdify(q, slope_expression, modules="numpy")
endpoint_change = float(evaluate(3.141592653589793) - evaluate(0.0))
for points in [101, 201, 401]:
    grid = np.linspace(0.0, 3.141592653589793, points)
    slopes = np.asarray(slope_function(grid), dtype=float)
    slopes = np.broadcast_to(slopes, grid.shape)
    integrated_change = np.trapezoid(slopes, grid)
    print(
        points,
        "integrated change:",
        integrated_change,
        "endpoint change:",
        endpoint_change,
    )

The trapezoidal rule joins neighbouring slope samples with straight segments and adds their signed areas. A negative contribution remains negative; replacing every slope by its magnitude would calculate a different quantity. Increasing the sample count provides a practical refinement check. If the values do not settle toward the endpoint difference, first inspect the interval, singularities, and array ordering.

Do not conclude that two agreeing calculations prove the model is physically complete. Both use the same selected relationship, so an omitted effect would be omitted from both. Their agreement is evidence that the numerical tools implement that relationship consistently. Comparing with an independent experiment or a more appropriate physical model is a separate test.

Keep this comparison in the notebook when you change the expression. It is particularly useful after adding a coefficient, changing a sign convention, or converting an input unit. A code cell that produces a smooth graph can conceal these mistakes; a second calculation of the same output change gives you another place to notice them.

Sources

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